Two Sets A and B Are as Under: A = {(a,b) ∈ R×R : |a−5| < 1 and |b−5| < 1}, B = {(a,b) ∈ R×R : 4(a−6)² + 9(b−5)² ≤ 36}. Then

JEE Main 2018. Previous Year Question.

Problem

Two sets $A$ and $B$ are as under: $A = \{(a,b)\in\mathbb{R}\times\mathbb{R} : |a-5|<1 \text{ and } |b-5|<1\}$; $B = \{(a,b)\in\mathbb{R}\times\mathbb{R} : 4(a-6)^2+9(b-5)^2\leq36\}$. Then:

Key insight. $A$ is an open square, and $B$ is a filled ellipse — a convex region. For a convex region, checking whether it contains a rectangle only requires checking the rectangle’s four corners: if every corner satisfies the defining inequality, every point inside does too, since the whole rectangle is a combination of its corners.

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Approach

Rewrite set $A$’s defining inequalities as a range for $a$ and $b$, identifying it as an open square. Rewrite set $B$’s inequality in standard ellipse form to identify its centre and axis lengths. Check whether all four corners of the square satisfy $B$’s inequality — since $B$ is convex, this confirms the entire square lies inside it.

Solution

Step 1 — Rewrite set A as a square

$$|a-5|<1 \implies 4<a<6, \qquad |b-5|<1 \implies 4<b<6$$

So $A$ is the open square with corners at $(4,4)$, $(4,6)$, $(6,4)$, $(6,6)$.

Step 2 — Rewrite set B in standard ellipse form

$$4(a-6)^2+9(b-5)^2\leq36 \implies \frac{(a-6)^2}{9}+\frac{(b-5)^2}{4}\leq1$$

This is a filled ellipse centred at $(6,5)$, with semi-axis $3$ in the $a$-direction and semi-axis $2$ in the $b$-direction.

Step 3 — Check each corner of the square against B’s inequality

At $(4,4)$: $\dfrac{(4-6)^2}{9}+\dfrac{(4-5)^2}{4} = \dfrac{4}{9}+\dfrac{1}{4} = \dfrac{16}{36}+\dfrac{9}{36}=\dfrac{25}{36}<1$ ✓

At $(4,6)$: $\dfrac{(4-6)^2}{9}+\dfrac{(6-5)^2}{4} = \dfrac{4}{9}+\dfrac{1}{4} = \dfrac{25}{36}<1$ ✓

At $(6,4)$: $\dfrac{(6-6)^2}{9}+\dfrac{(4-5)^2}{4} = 0+\dfrac{1}{4}<1$ ✓

At $(6,6)$: $\dfrac{(6-6)^2}{9}+\dfrac{(6-5)^2}{4} = 0+\dfrac{1}{4}<1$ ✓

Step 4 — Use convexity to conclude for the whole square

Since $B$’s defining region is convex (an ellipse’s interior), and its defining expression achieves its maximum over a rectangle at one of the rectangle’s corners, all four corners satisfying the inequality strictly means every point of the (closed) square also satisfies it strictly. So the entire square $A$ lies inside $B$.

Step 5 — Conclude the subset relation

$$A \subset B$$

(This is strict containment, since $B$ contains many points, such as points near its own boundary, that lie outside the square $A$.)

Answer

Common mistakes

• Trying to check infinitely many points of the square individually. Since $B$’s region is convex, checking only the four corners is sufficient — there’s no need to test interior points or the edges separately.
• Confusing which set is the square and which is the ellipse. $A$’s defining condition (two absolute value inequalities) describes a square; $B$’s condition (a sum of squared terms $\leq$ a constant) describes an ellipse — mixing these up would reverse the entire analysis.

Practise next

• Let $P = \{(x,y): |x-3|<2, |y-4|<2\}$ and $Q = \{(x,y): (x-4)^2/16+(y-4)^2/9\leq1\}$. Determine the relationship between $P$ and $Q$, using the same corner-checking method.